Show that Brownian motion on the unit circle is exponentially ergodic and has the uniform measure as its invariant distribution. My search results keep bring up planar Brownian motion on the unit disk. However, I am specifically referring to 
$e^{jW_{t}} = [\cos(W_t),\sin(W_t)]^{T}$
where $W_t$ is Brownian motion.
I am at a loss here, so any advice to point me in the right direction would be greatly appreciated. Thanks in advance.
 A: 1. The uniform distribution is invariant
Let $\theta_0$ have a uniform distribution in the circle, so that its pdf is $f(\theta)= \frac 1{2\pi}$ for $\theta \in [0,2\pi]$ and $0$ otherwise.
If $\theta_0$ moves according to the law of Brownian motion during a time interval $[0,t]$, then its new distribution over the reals can be calculated by applying the Brownian transition function to the initial density $f$. 
Note that arguments $\theta$ and $\theta + 2k\pi$ represent the same point in the circle, so in order to get the probability density $f_t$ over the circle their probability densities should be added . If we write $p_t(x,y)$ for the standard Brownian transition density 
$$p_t(x,y) = \frac{1}{\sqrt{2\pi t}}e^{-\frac{(y-x)^2}{2t}}$$
 between reals $x$ and $y$, then the density $q_t(\phi, \theta)$ between points in the circle would read $$q_t(\phi, \theta) = 
\cases{
    \sum_{k\in\mathbb Z} p_t(\phi, \theta + 2k\pi) & \text{if $\phi, \theta\in [0,2\pi]$,} \\ 
    0 & \text{otherwise.}
    }$$
Using this new transition density, the distribution of the point at time $t$ is
$$\begin{align}
f_t(\theta) 
&= \int_{0}^{2\pi}q_t(\phi,\theta)f(\phi) \, d\phi\\
&= \sum_{k\in\mathbb Z}\int_{0}^{2\pi}p_t(\phi,\theta + 2k\pi)f(\phi) \, d\phi\\ 
&= \sum_{k\in\mathbb Z} \int_{0}^{2\pi} p_t(\phi - 2k\pi,\theta) \frac 1{2\pi} \, d\phi\\
&= \frac 1{2\pi}\sum_{k\in\mathbb Z} \int_{-2k\pi}^{-2(k-1)\pi} p_t(\phi,\theta) \, d\phi\\
&= \frac 1{2\pi} \int_{-\infty}^{\infty} p_t(\phi,\theta) \, d\phi = \frac 1{2\pi} = f(\theta).\\
\end{align}$$
(Changes in the order of summation are justified by the monotone convergence theorem, as the quantities are positive.)
In other words, the distribution doesn't change, as we wished to show.
2. Exponential ergodicity 
Now we want to establish the exponential ergodicity of Brownian motion in the circle towards the uniform distribution. Writing $Q^t(x,A)$ for its transition function and $\mu$ for the uniform probability measure, this means that$^1$ $$\sup_{A\in\mathcal B[0,2\pi]}|Q^t(x,A)-\mu(A)| \le M(x)\rho^{t} $$ for some positive $\rho < 1$, finite $M(x)$. Here I used the definition of the total variation distance between probability measures.
We will actually prove that the pdf of $\theta_t = W_t \bmod 2\pi$ converges to the pdf of the uniform distribution over time. Indeed, the probability density of $\theta_t$ starting at $0$ is 
\begin{align}
g_t(\phi)=\sum_{k\in \mathbb Z} p_t(0,\phi+2k\pi)
&= \sum_{k\in \mathbb Z} \frac 1 {\sqrt{2\pi t}}e^{-\frac 1{2t}(\phi+2k\pi)^2}\\
&= \frac 1 {\sqrt{2\pi t}}e^{-\phi^2/2t}\sum_{k\in \mathbb Z} e^{-2k\pi\phi/t}e^{ -2k^2\pi^2/t}.
\end{align}
We want to know the limit of this last expression as $t\to\infty$. Note that the terms of the series approach $1$ and it's difficult to see clearly how that compares to the decreasing factor outside the sum. Still, following the approach in this post, we can express it in terms of Jacobi's theta function $$\vartheta_3(z,q) = \sum_{k\in\mathbb Z}q^{k^2}e^{2kiz}$$ and apply Jacobi's imaginary transformation (more on that below) to get a more tractable expression. 
Applying the transformation involves introducing a new variable $\tau$ defined by $q=e^{i\pi\tau}$. The same function expressed in terms of this variable reads $$\vartheta_3(z\mid \tau) = \sum_{k\in\mathbb Z}e^{i\pi\tau k^2}e^{2kiz}.$$ In our case we have 
$$z =\frac{\phi\pi i}{t},\quad q = e^{-2\pi^2/t},\quad \tau=\frac{2\pi i}{t}$$ 
and using this notation our expression becomes
$$g_t(\phi) = \frac 1 {\sqrt{2\pi t}}e^{-\phi^2/2t} \vartheta_3(z\mid \tau).$$
Now we can apply the imaginary transformation $(6)$ in this link, and get the equivalent form
$$\frac 1 {\sqrt{2\pi t}}e^{-\phi^2/2t} \sqrt{\frac{t}{2\pi}} e^{-iz^2/\tau\pi} \vartheta_3\Bigr(-\frac{z}{\tau} \Bigr | \Bigl. -\frac{1}{\tau}\Bigl) = \frac 1{2\pi} \vartheta_3\Bigr(-\frac{\phi}{2} \Bigr | \Bigl. \frac{it}{2\pi}\Bigl) =  \frac 1{2\pi} \vartheta_3(z', q'),$$
with $z' = -\phi/2,\,\, q'= e^{-t/2}$. Finally, notice that $q'$ approaches $ 0$ as $t\to\infty$, therefore only the constant term survives in the series $\vartheta_3(z', q')$. Thus, we have pointwise convergence of the density $g_t(\phi) \rightarrow \frac 1{2\pi}=f(\theta)$ for all $\phi\in[0,2\pi]$. Note that the same argument holds for the process started at another point $\theta_0$ if we replace $\phi$ by  $\phi-\theta_0 $.
This establishes convergence in distribution of the Brownian motion on the circle to the uniform distribution. Moreover,for any borel subset $A$ of $[0,2\pi]$ we have
$$|Q^t(\theta_0,A)-\mu(A)| 
= \left |\int_A Q^t(\theta_0, d\phi) -\int_A f(\theta)\,d\phi\right| 
\le  \int_A |g_t(\phi-\theta_0)- f(\theta)|\,d\phi $$
$$= \int_A \frac 1{2\pi}|\vartheta_3(z', q')- 1|\,d\phi 
= \int_A \frac 1{2\pi}\left|\sum_{k\ne 0}e^{-tk^2/2} e^{-ki(\phi-\theta_0)}\right|\,d\phi
\le \int_A \frac 1{2\pi}\sum_{k\ne 0 }e^{-tk^2/2}\,d\phi$$
$$ \le 2\sum_{k\ge 1}e^{-tk^2/2} \le 2\sum_{k\ge 1}e^{-tk/2} = 2\frac{e^{-t/2}}{1-e^{-t/2}} \le 4e^{-t/2} $$
for $t\ge \log 4$, so that the process is exponentially ergodic with $M=4, \rho=e^{-1/2}$, as we wished to show.



*

*Ergodicity definition taken from this article
